Indirect Proof

Date: 02/14/98 at 21:06:02
From: Jamie Hanson
Subject: Indirect  Proofs

Prove that sqrt 2 is an irrational number.

I have no ideas.  Can you help?

Date: 02/17/98 at 19:14:02
From: Doctor Wolf
Subject: Re: Indirect  Proofs

Hi Jamie,
Let's say that sqrt(2) was rational; that is, it could be written as
a fraction a/b in REDUCED form where both a and b are positive 
integers.

Reduced form would imply that the integers a and b could not both be
even, otherwise we could reduce the fraction even further by dividing
both a and b by 2 (example: 8/12 = 4/6 = 2/3).

Since a/b = sqrt(2) we would have    a^2       2
                                  --------- = --- (squaring both
                                     b^2       1   sides of equation)

It gets a little tricky here ... so let's go slowly!  There is a 
property of numbers I will need (not too hard to prove): Only even
integers will produce even integers when squared.

Back to the problem. If I cross multiply in the equation above, I'll
have a^2 = 2b^2.  Well, 2b^2 is certainly an even integer (since it 
has 2 as a factor), and so both a^2 AND a (using number property) must 
be even.  Okay ... so we've shown a is an even integer ... now what?
What if we could show that b is also even ... a contradiction!

Since a is an even integer, we can write it as 2c where c is just a 
divided by 2 (for example: 10 = 2*5). Now let's rewrite our original
equation replacing a by 2c:

     (2c)^2    2       4c^2    2
     ------  = -  or   ---- = ---  or (after cross multiplying)
      b^2      1        b^2    1

      4c^2 = 2b^2.  After dividing both sides by 2, 2c^2 = b^2.

What does this say (odd or even integer) about 2c^2, b^2, and b? And 
why is this a contradiction of our original assumption about a/b?

Can you take it from here Jamie ...(Remember to use number property.)

P.S.: There are lots of questions like yours and their solutions in
the Dr. Math Archives ... you might want to check them out.

-Doctor Wolf,  The Math Forum
 Check out our Web site   http://mathforum.org/dr.math/